Chapter 10 2 Page 2 8. The subtended angle in radians is the size of the object divided by the distance to the object. A pencil with a diameter of 6 mm will block out the Moon if it is held about 60 cm from the eye. For the angle subtended we have θMoon= Dpencil/rpencil˜ (0.6 cm)/(60 cm) ˜ 0.01 rad. We estimate the diameter of the Moon from Moon= DMoon/rMoon; 0.01 rad = DMoon/(3.8 × 105km), which gives DMoon˜ 4 × 103km. 9. (a) ω= (2500 rev/min)(2p rad/rev)/(60 s/min) = 262 rad/s. (b) The linear speed of the point on the edge is the tangential speed: v = r= (0.175 m)(262 rad/s) = 46 m/s. Because the speed is constant, the tangential acceleration is zero. There will be a radial acceleration: aR= 2R= (262 rad/s)2(0.175 m) = 1.2 × 104m/s2radial. 10. (a) The angular speed of the merry-go-round is = (1 rev)(2p rad/rev)/(4.0 s) = 1.57 rad/s. The linear speed of the child is the tangential speed: v = r= (1.2 m)(1.57 rad/s) = 1.9 m/s. (b) The child will have a radial acceleration: aR= 2R= (1.57 rad/s)2(1.2 m) = 3.0 m/s2radial. 11. In each revolution the ball rolls a distance equal to its circumference, so we have L= N(pD); 3.5 m = (15.0)pD, which gives D= 0.074 m = 7.4 cm.

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